10-F1127PA025SC-L167E09 Output Inverter Application flow 7PACK 1 1200 V / 25 A General conditions 3phase SPWM V GEon = 15 V V GEoff = -15 V R gon = 16 Ω R goff = 16 Ω Figure 1 IGBT Figure 2 Typical average static loss as a function of output current P loss = f(I out) 60 Ploss (W) 80 Ploss (W) FWD Typical average static loss as a function of output current P loss = f(I out) Mi*cosfi = 1 70 Mi*cosf i= -1 50 60 40 50 30 40 30 20 20 10 10 Mi*cosfi = 1 Mi*cosfi = -1 0 0 0 5 10 15 20 25 30 35 40 45 0 50 5 10 15 20 25 30 35 40 Iout (A) At Tj = 150 At Tj = °C Mi*cosφ from -1 to 1 in steps of 0,2 50 150 °C Mi*cosφ from -1 to 1 in steps of 0,2 Figure 3 IGBT Typical average switching loss as a function of output current Figure 4 FWD Typical average switching loss as a function of output current P loss = f(I out) 80,0 Ploss (W) Ploss (W) 45 Iout (A) 70,0 fsw = 16kHz P loss = f(I out) 25,0 20,0 60,0 fsw = 16kHz 50,0 15,0 40,0 10,0 30,0 20,0 5,0 10,0 fsw = 2kHz fsw = 2kHz 0,0 0,0 0 At Tj = 5 150 10 15 20 25 30 35 40 0 45 50 Iout (A) At Tj = °C DC link = 600 V f sw from 2 kHz to 16 kHz in steps of factor 2 copyright Vincotech 5 150 10 15 20 25 30 35 40 45 50 Iout (A) °C DC link = 600 V f sw from 2 kHz to 16 kHz in steps of factor 2 1 27 Jul. 2015 / Revision 1 10-F1127PA025SC-L167E09 Output Inverter Application flow 7PACK 1 Figure 5 Phase Figure 6 Typical available 50Hz output current as a function Mi*cosφ I out = f(Mi*cos φ ) Iout (A) Iout (A) Phase Typical available 50Hz output current as a function of switching frequency I out = f(f sw) 40 Th = 60°C 35 30 1200 V / 25 A 40 Th = 60°C 35 30 Th = 100°C 25 25 20 20 15 15 10 10 5 5 Th = 100°C 0 0 -1,0 -0,8 At Tj = -0,6 150 -0,4 -0,2 0,0 0,2 0,4 0,6 1 0,8 1,0 Mi*cos φ 10 100 fsw (kHz) At Tj = °C DC link = 600 V f sw = 4 kHz T h from 60 °C to 100 °C in steps of 5 °C 150 °C DC link = 600 V Mi*cos φ =0,8 T h from 60 °C to 100 °C in steps of 5 °C Figure 7 Phase Figure 8 Typical available 50Hz output current as a function of Mi*cos φ and switching frequency I out = f(f sw, Mi*cos φ ) Phase Typical available 0Hz output current as a function of switching frequency I outpeak = f(f sw) Iout (Apeak) -1,00 -0,80 Iout (A) 40 35 -0,60 30 35,0-40,0 -0,40 30,0-35,0 Mi*cosfi 0,00 20,0-25,0 Th = 60°C 25 -0,20 25,0-30,0 20 0,20 15 15,0-20,0 Th = 100°C 0,40 10 10,0-15,0 0,60 5 0,80 0 1,00 1 At Tj = 2 4 fsw8 16 32 1 64 10 100 fsw (kHz) 150 °C At Tj = DC link = 600 Th = 80 V °C DC link = 600 V T h from 60 °C to 100 °C in steps of 5 °C Mi = copyright Vincotech 2 150 °C 0 27 Jul. 2015 / Revision 1 10-F1127PA025SC-L167E09 Output Inverter Application flow 7PACK 1 Figure 9 Inverter Figure 10 Inverter Typical efficiency as a function of output power efficiency=f(Pout) efficiency (%) Typical available peak output power as a function of heatsink temperature P out=f(T h) Pout (kW) 1200 V / 25 A 18,0 2kHz 16,0 100,0 99,0 2kHz 98,0 14,0 97,0 12,0 96,0 10,0 16kHz 95,0 16kHz 8,0 94,0 6,0 93,0 4,0 92,0 2,0 91,0 0,0 60 65 70 75 80 85 90 At Tj = 150 DC link = Mi = cos φ= f sw from 600 V 1 0,80 2 kHz to 16 kHz in steps of factor 2 90,0 95 100 Th (oC) 0,0 10,0 15,0 20,0 25,0 Pout (kW) °C Figure 11 5,0 At Tj = 150 DC link = Mi = cos φ= f sw from 600 V 1 0,80 2 kHz to 16 kHz in steps of factor 2 °C Inverter Overload (%) Typical available overload factor as a function of motor power and switching frequency P peak / P nom=f(P nom,fsw) 400 350 300 250 200 150 Switching frequency (kHz) 100 Motor nominal power (HP/kW) 5,00 / 3,68 7,50 / 5,52 10,00 / 7,36 15,00 / 11,03 20,00 / 14,71 25,00 / 18,39 1 374 250 187 125 0 0 2 374 250 187 125 0 0 4 374 250 187 125 0 0 8 374 250 187 125 0 0 16 294 196 147 0 0 0 At Tj = 150 DC link = Mi = cos φ= f sw from Th = 600 V 1 0,8 1 kHz to 16kHz in steps of factor 2 80 °C °C Motor eff =0,85 copyright Vincotech 3 27 Jul. 2015 / Revision 1